What is (\partial ^2)/(\partial x\partial y)(y^4cos(2x)) ?

The answer to (\partial ^2)/(\partial x\partial y)(y^4cos(2x)) is -8y^3sin(2x)

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(\partial ^2)/(\partial x\partial y)(y^4cos(2x))

\frac{\partial ^{2}}{\partial x \partial y} (y^{4} cos (2 x))

- 8 y^{3} sin (2 x)

Solution steps

\frac{\partial ^{2}}{\partial x \partial y} (y^{4} cos (2 x))

\frac{\partial}{\partial x} (y^{4} cos (2 x)) = - 2 y^{4} sin (2 x)

= \frac{\partial}{\partial y} (- 2 y^{4} sin (2 x))

\frac{\partial}{\partial y} (- 2 y^{4} sin (2 x)) = - 8 y^{3} sin (2 x)

= - 8 y^{3} sin (2 x)

\frac{\partial}{\partial y} (16 - x^{2} - 4 y)

y^{^{''}} - y = sin (2 t)

f (x) = \frac{e ^{x}}{x + 3}

\frac{( s + 5 )}{( s ^{2} + 4 s + 13 )}

\frac{d}{d x} (x^{2} (1 - x)^{3})

What is (\partial ^2)/(\partial x\partial y)(y^4cos(2x)) ?
The answer to (\partial ^2)/(\partial x\partial y)(y^4cos(2x)) is -8y^3sin(2x)